Given a binary relation $\mathcal R$ over a set $A$,then the reflexive closure of $\mathcal R$ on $A$ denoted by $\mathcal S$ is the smallest reflexive relation on $A$ containing $\mathcal R$.
Equivalently it's the least reflexive relation on $A$ that is a superset of $\mathcal R$.
Reflexive closure is explicitly given by:$$\mathcal S=\text{id}_A \cup\mathcal R$$
Since $\mathcal S$ is reflexive,hence by definition of reflexivity $\text{id}_A \subseteq \mathcal S$,on the other hand since it contains $\mathcal R$,implies that:
$$\text{id}_A \cup \mathcal R \subseteq \mathcal S$$
Therefore $\mathcal S $ can be written as: $$\mathcal S=\text{id}_A \cup \mathcal R\cup B$$
It's left to show that $B=\varnothing$,assume for the sake of contradiction $B \ne \varnothing$,then there is another reflexive closure $\mathcal S '$ with $B=\varnothing$ which is indeed $\mathcal S'=\text{id}_A \cup \mathcal R$ ,from here it's seen that $\mathcal S' \subset \mathcal S$,contradicts the fact that $\mathcal S$ is the smallest such reflexive relation on $A$ containing $ \mathcal R$.$\blacksquare$
All I tried to show,was the validity of such characterization of the reflexive closure.
However I'm not sure if my arguments are right,it would be highly appreciated if someone check them.
You can avoid the use of contradiction (which is a bit confusing).
It is enough to observe two things:
This justifies the conclusion that $\text{id}_A \cup\mathcal R$ is the reflexive closure of $\mathcal R$.
Both bullets are evident.
You use the notation $\text{id}_A$ which is dubious (as commented).
Often notations $\Delta(A)$ or $\Delta_A$ are used in this context.
This subset of $A\times A$ is called the "diagonal".